Weak compactness in the James space and its dual

One of the things which you can do is: to combine the following theorem of James [James, Robert C. Weakly compact sets. Trans. Amer. Math. Soc. 113 1964 129–140]: A weakly closed subset $C$ of a Banach space $B$ is weakly compact if and only if each member of $B^*$ attains a maximum on $C$ with the description of the dual space of $J$ and $J^*$.

Another thing: analyze when, say, the weak$^*$-limit of a sequence in $J$, considered as a sequence in $J^{**}$, is outside $J$. Do the same for the pair $J^*$ and $J^{***}$

I think that one can definitely get something on these lines. Possibly this has already been done somewhere, but it could be difficult to find.

I'm not sure this deserves to be posted as an "answer," but it is far too long for a comment. But since there is already an answer posted, I do not believe this will detract from your question getting attention. Anyway...

First, note that $J$ has a unique (up to equivalence) spreading basis, which we shall call "the" spreading basis for $J$. As every seminormalized weakly null sequence in $J$ contains a subsequence equivalent to the $\ell_2$ basis, the spreading basis cannot be weakly null. This was all established here: http://www.mscand.dk/article/viewFile/11904/9920

From the same paper comes the following (as part (b) from Theorem 2.1).

Theorem. Let $(z_n)_{n=1}^\infty$ be a seminormalized sequence in $J$ having no weak cluster point. Then there is a subsequence $(z_{n_k})_{k=1}^\infty$ equivalent to the spreading basis of $J$, such that its closed linear span $[z_{n_k}]_{k=1}^\infty$ is complemented in $J$.

Corollary. Let $C\subset J$ be a subset which is weakly closed. Then the following are equivalent.

(i) $C$ is not weakly compact.

(ii) $C$ contains a sequence $(c_n)_{n=1}^\infty$ equivalent to the spreading basis of $J$ such that its closed linear span $[c_n]_{n=1}^\infty$ is complemented in $J$.

(iii) $C$ contains a basic sequence $(c_n)_{n=1}^\infty$ which is not weakly null.

Proof. (i) $\Rightarrow$ (ii): By Eberlein-Smulian we can find $(c_n)_{n=1}^\infty\subset C$ with no weak cluster point in $C$, and hence (by weak closure of $C$) no weak cluster point in $J$. Now apply the above theorem to obtain a subsequence equivalent to $J$ and complemented in $J$.

(ii) $\Rightarrow$ (iii): This follows from the fact that the spreading basis is not weakly null.

(iii) $\Rightarrow$ (i): By passing to a subsequence if necessary we may assume that the weak closure of $(c_n)_{n=1}^\infty$ (as a set) fails to contain zero. Hence, by a well-known criterion of Kadets-Pelczynski (e.g. Theorem 1.5.6 in the Albiac/Kalton book here), $(c_n)_{n=1}^\infty$ is not relatively weakly compact, and by Eberlein-Smulian, has no weakly convergent subsequence. Thus $C$ is not weakly compact. $\square$

Note that the equivalence (i) $\Leftrightarrow$ (iii) holds for arbitrary Banach space in place of $J$. So, the only special thing here is (ii).